3.45 \(\int \frac {x (a+b \tanh ^{-1}(c \sqrt {x}))}{d+e x} \, dx\)

Optimal. Leaf size=374 \[ \frac {2 d \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{e^2}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^2 e}-\frac {b d \text {Li}_2\left (1-\frac {2}{\sqrt {x} c+1}\right )}{e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (\sqrt {-d} c+\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 e^2}+\frac {b \sqrt {x}}{c e} \]

[Out]

-b*arctanh(c*x^(1/2))/c^2/e+x*(a+b*arctanh(c*x^(1/2)))/e+2*d*(a+b*arctanh(c*x^(1/2)))*ln(2/(1+c*x^(1/2)))/e^2-
d*(a+b*arctanh(c*x^(1/2)))*ln(2*c*((-d)^(1/2)-e^(1/2)*x^(1/2))/(c*(-d)^(1/2)-e^(1/2))/(1+c*x^(1/2)))/e^2-d*(a+
b*arctanh(c*x^(1/2)))*ln(2*c*((-d)^(1/2)+e^(1/2)*x^(1/2))/(c*(-d)^(1/2)+e^(1/2))/(1+c*x^(1/2)))/e^2-b*d*polylo
g(2,1-2/(1+c*x^(1/2)))/e^2+1/2*b*d*polylog(2,1-2*c*((-d)^(1/2)-e^(1/2)*x^(1/2))/(c*(-d)^(1/2)-e^(1/2))/(1+c*x^
(1/2)))/e^2+1/2*b*d*polylog(2,1-2*c*((-d)^(1/2)+e^(1/2)*x^(1/2))/(c*(-d)^(1/2)+e^(1/2))/(1+c*x^(1/2)))/e^2+b*x
^(1/2)/c/e

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Rubi [A]  time = 0.48, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {43, 5980, 5916, 321, 206, 6044, 5920, 2402, 2315, 2447} \[ -\frac {b d \text {PolyLog}\left (2,1-\frac {2}{c \sqrt {x}+1}\right )}{e^2}+\frac {b d \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{2 e^2}+\frac {b d \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{2 e^2}+\frac {2 d \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{e^2}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^2 e}+\frac {b \sqrt {x}}{c e} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcTanh[c*Sqrt[x]]))/(d + e*x),x]

[Out]

(b*Sqrt[x])/(c*e) - (b*ArcTanh[c*Sqrt[x]])/(c^2*e) + (x*(a + b*ArcTanh[c*Sqrt[x]]))/e + (2*d*(a + b*ArcTanh[c*
Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/e^2 - (d*(a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/(
(c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/e^2 - (d*(a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*Sqr
t[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/e^2 - (b*d*PolyLog[2, 1 - 2/(1 + c*Sqrt[x])])/e^2 + (b*d*Pol
yLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2) + (b*d*PolyL
og[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5920

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTanh[c*x])
^p)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 6044

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && (GtQ[q, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d+e x} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{d+e x^2} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 \operatorname {Subst}\left (\int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx,x,\sqrt {x}\right )}{e}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{d+e x^2} \, dx,x,\sqrt {x}\right )}{e}\\ &=\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}-\frac {(b c) \operatorname {Subst}\left (\int \frac {x^2}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e}-\frac {(2 d) \operatorname {Subst}\left (\int \left (-\frac {a+b \tanh ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {a+b \tanh ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx,x,\sqrt {x}\right )}{e}\\ &=\frac {b \sqrt {x}}{c e}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}+\frac {d \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{e^{3/2}}-\frac {d \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{e^{3/2}}-\frac {b \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{c e}\\ &=\frac {b \sqrt {x}}{c e}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^2 e}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}+\frac {2 d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}-2 \frac {(b c d) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e^2}+\frac {(b c d) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) (1+c x)}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e^2}+\frac {(b c d) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) (1+c x)}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e^2}\\ &=\frac {b \sqrt {x}}{c e}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^2 e}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}+\frac {2 d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e^2}-2 \frac {(b d) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c \sqrt {x}}\right )}{e^2}\\ &=\frac {b \sqrt {x}}{c e}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{c^2 e}+\frac {x \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}+\frac {2 d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e^2}-\frac {b d \text {Li}_2\left (1-\frac {2}{1+c \sqrt {x}}\right )}{e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e^2}+\frac {b d \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e^2}\\ \end {align*}

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Mathematica [A]  time = 1.46, size = 337, normalized size = 0.90 \[ \frac {-2 a d \log (d+e x)+2 a e x+\frac {2 b \left (\tanh ^{-1}\left (c \sqrt {x}\right ) \left (2 c^2 d \log \left (e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}+1\right )+c^2 e x-e\right )-c^2 d \text {Li}_2\left (-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+c^2 d \tanh ^{-1}\left (c \sqrt {x}\right )^2+c e \sqrt {x}\right )}{c^2}-b d \left (\text {Li}_2\left (-\frac {\left (d c^2+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{d c^2-2 \sqrt {-d} \sqrt {e} c-e}\right )+\text {Li}_2\left (-\frac {\left (d c^2+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{d c^2+2 \sqrt {-d} \sqrt {e} c-e}\right )+2 \tanh ^{-1}\left (c \sqrt {x}\right ) \left (\log \left (\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d-2 c \sqrt {-d} \sqrt {e}-e}+1\right )+\log \left (\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d+2 c \sqrt {-d} \sqrt {e}-e}+1\right )-\tanh ^{-1}\left (c \sqrt {x}\right )\right )\right )}{2 e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(a + b*ArcTanh[c*Sqrt[x]]))/(d + e*x),x]

[Out]

(2*a*e*x - 2*a*d*Log[d + e*x] + (2*b*(c*e*Sqrt[x] + c^2*d*ArcTanh[c*Sqrt[x]]^2 + ArcTanh[c*Sqrt[x]]*(-e + c^2*
e*x + 2*c^2*d*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])]) - c^2*d*PolyLog[2, -E^(-2*ArcTanh[c*Sqrt[x]])]))/c^2 - b*d*(
2*ArcTanh[c*Sqrt[x]]*(-ArcTanh[c*Sqrt[x]] + Log[1 + ((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d - 2*c*Sqrt[-
d]*Sqrt[e] - e)] + Log[1 + ((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d + 2*c*Sqrt[-d]*Sqrt[e] - e)]) + PolyL
og[2, -(((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d - 2*c*Sqrt[-d]*Sqrt[e] - e))] + PolyLog[2, -(((c^2*d + e
)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d + 2*c*Sqrt[-d]*Sqrt[e] - e))]))/(2*e^2)

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \operatorname {artanh}\left (c \sqrt {x}\right ) + a x}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*x*arctanh(c*sqrt(x)) + a*x)/(e*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )} x}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)*x/(e*x + d), x)

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maple [A]  time = 0.06, size = 539, normalized size = 1.44 \[ \frac {a x}{e}-\frac {a d \ln \left (c^{2} e x +c^{2} d \right )}{e^{2}}+\frac {b \arctanh \left (c \sqrt {x}\right ) x}{e}-\frac {b \arctanh \left (c \sqrt {x}\right ) d \ln \left (c^{2} e x +c^{2} d \right )}{e^{2}}-\frac {b d \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 e^{2}}+\frac {b d \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 e^{2}}+\frac {b d \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 e^{2}}+\frac {b d \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 e^{2}}+\frac {b d \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 e^{2}}+\frac {b d \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 e^{2}}-\frac {b d \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 e^{2}}-\frac {b d \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 e^{2}}-\frac {b d \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 e^{2}}-\frac {b d \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 e^{2}}+\frac {b \sqrt {x}}{c e}+\frac {b \ln \left (c \sqrt {x}-1\right )}{2 c^{2} e}-\frac {b \ln \left (1+c \sqrt {x}\right )}{2 c^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x)

[Out]

a*x/e-a/e^2*d*ln(c^2*e*x+c^2*d)+b*arctanh(c*x^(1/2))*x/e-b*arctanh(c*x^(1/2))/e^2*d*ln(c^2*e*x+c^2*d)-1/2*b/e^
2*d*ln(c*x^(1/2)-1)*ln(c^2*e*x+c^2*d)+1/2*b/e^2*d*ln(c*x^(1/2)-1)*ln((c*(-d*e)^(1/2)-e*(c*x^(1/2)-1)-e)/(c*(-d
*e)^(1/2)-e))+1/2*b/e^2*d*ln(c*x^(1/2)-1)*ln((c*(-d*e)^(1/2)+e*(c*x^(1/2)-1)+e)/(c*(-d*e)^(1/2)+e))+1/2*b/e^2*
d*dilog((c*(-d*e)^(1/2)-e*(c*x^(1/2)-1)-e)/(c*(-d*e)^(1/2)-e))+1/2*b/e^2*d*dilog((c*(-d*e)^(1/2)+e*(c*x^(1/2)-
1)+e)/(c*(-d*e)^(1/2)+e))+1/2*b/e^2*d*ln(1+c*x^(1/2))*ln(c^2*e*x+c^2*d)-1/2*b/e^2*d*ln(1+c*x^(1/2))*ln((c*(-d*
e)^(1/2)-e*(1+c*x^(1/2))+e)/(c*(-d*e)^(1/2)+e))-1/2*b/e^2*d*ln(1+c*x^(1/2))*ln((c*(-d*e)^(1/2)+e*(1+c*x^(1/2))
-e)/(c*(-d*e)^(1/2)-e))-1/2*b/e^2*d*dilog((c*(-d*e)^(1/2)-e*(1+c*x^(1/2))+e)/(c*(-d*e)^(1/2)+e))-1/2*b/e^2*d*d
ilog((c*(-d*e)^(1/2)+e*(1+c*x^(1/2))-e)/(c*(-d*e)^(1/2)-e))+b*x^(1/2)/c/e+1/2/c^2*b*ln(c*x^(1/2)-1)/e-1/2/c^2*
b*ln(1+c*x^(1/2))/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + b \int \frac {x \log \left (c \sqrt {x} + 1\right )}{2 \, {\left (e x + d\right )}}\,{d x} - b \int \frac {x \log \left (-c \sqrt {x} + 1\right )}{2 \, {\left (e x + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x, algorithm="maxima")

[Out]

a*(x/e - d*log(e*x + d)/e^2) + b*integrate(1/2*x*log(c*sqrt(x) + 1)/(e*x + d), x) - b*integrate(1/2*x*log(-c*s
qrt(x) + 1)/(e*x + d), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x\,\left (a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*atanh(c*x^(1/2))))/(d + e*x),x)

[Out]

int((x*(a + b*atanh(c*x^(1/2))))/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x**(1/2)))/(e*x+d),x)

[Out]

Integral(x*(a + b*atanh(c*sqrt(x)))/(d + e*x), x)

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